The surface area of a solid sphere is increased by 21% without changing its shape. Find the percentage increase in its :

(i) radius

(ii) volume

Let original radius be r units and new radius be R units.

Given,

Surface area of a solid sphere is increased by 21% without changing its shape.

$\therefore 4πR^2 = 4πr^2 + \dfrac{21}{100} \times 4πr^2 \\[1em] \Rightarrow 4πR^2 = 4π(r^2 + \dfrac{21}{100}r^2) \\[1em] \Rightarrow R^2 = r^2 + \dfrac{21}{100}r^2 \\[1em] \Rightarrow R^2 = \dfrac{121}{100}r^2 \\[1em] \Rightarrow R = \sqrt{\dfrac{121}{100}r^2} \\[1em] \Rightarrow R = \dfrac{11}{10}r.$

By formula,

$\text{Percentage increase in radius} = \dfrac{R - r}{r} \times 100 \\[1em] = \dfrac{\dfrac{11}{10}r - r}{r} \times 100 \\[1em] = \dfrac{\dfrac{r}{10}}{r} \times 100 \\[1em] = \dfrac{1}{10} \times 100 \\[1em] = 10\%.$

Hence, percentage increase in radius = 10%.

(ii) Let original volume be v and new volume be V.

By formula,

$\text{Percentage increase in volume} = \dfrac{V - v}{v} \times 100 \\[1em] = \dfrac{\dfrac{4}{3}πR^3 - \dfrac{4}{3}πr^3}{\dfrac{4}{3}πr^3} \times 100\\[1em] = \dfrac{\dfrac{4}{3}π(R^3 - r^3)}{\dfrac{4}{3}πr^3} \times 100\\[1em] = \dfrac{R^3 - r^3}{r^3} \times 100 \\[1em] = \dfrac{\Big(\dfrac{11}{10}r\Big)^3 - r^3}{r^3} \times 100 \\[1em] = \dfrac{\dfrac{1331}{1000} - 1}{1} \times 100 \\[1em] = \dfrac{\dfrac{1331 - 1000}{1000}}{1} \times 100 \\[1em] = \dfrac{331}{1000}\times 100 \\[1em] = 33.1\%.$

Hence, percentage increase in volume = 33.1%.